Sandy has $5,000 invested in three accounts. one account bears 2% annual interest, another 3% and other 5%. she has four times the amount invested at 5% as she does at 2%, and she earned $202.50 for the year. write and solve a system of equations to find out how much money Sandy has in each account.

Question

Sandy has $5,000 invested in three accounts. one account bears 2% annual interest, another 3% and other 5%. she has four times the amount invested at 5% as she does at 2%, and she earned $202.50 for the year. write and solve a system of equations to find out how much money Sandy has in each account.

2%=___________
3%=___________
5%=___________

Answer 1

I will get you going with the definition

amount invested at 2% ---- x
amount invested at 5% ---- 4x
amount invested at 3% ---- 5000 - 5x

return from the 2% account = .02x
return from the 5% account = .05(4x)
return from the 3% account = .....

Now their sum = 202.50

make up the equation, and solve

Answer 2

yes but i have to use the linear systems in three variables.

so like this .02x+.03y+.05z=5,000

Answer 3

Invested $X at 2%.

Invested $4x at 5%.
Invested $(5000 - 5x) at 3%.

I = Po1*r1*t + Po2*r2*t + Po3*r3*t = 202.50.
x*0.02*1 + 4x*0.05*1 + (5000-5x)*0.03*1 = 202.50,
X = ?
4x = ?
5000-5x = ?

Answer 4

Let's assign variables to represent the amounts of money Sandy has invested in each account.

Let's say:
x = amount invested at 2% interest
y = amount invested at 3% interest
z = amount invested at 5% interest

We can begin by setting up three equations based on the information given:

1) Sandy has $5,000 invested in total: x + y + z = 5000

2) Sandy has four times the amount invested at 5% as she does at 2%: z = 4x

3) Sandy earned $202.50 for the year: (0.02x) + (0.03y) + (0.05z) = 202.50

To solve this system of equations, we can use the substitution method.

First, let's substitute z in equation 1 with the value from equation 2:

x + y + 4x = 5000
5x + y = 5000

Now, let's substitute z in equation 3 with the value from equation 2:

(0.02x) + (0.03y) + (0.05 * 4x) = 202.50
0.02x + 0.03y + 0.20x = 202.50
0.22x + 0.03y = 202.50

We now have a system of two equations with two variables:

5x + y = 5000 (Equation 4)
0.22x + 0.03y = 202.50 (Equation 5)

To solve this system, we can either solve equation 4 for y and substitute it into equation 5, or use the elimination method. Let's use the elimination method to solve this system.

Multiply Equation 5 by 100 to eliminate the decimal:

22x + 3y = 20250 (Equation 6)

Now, multiply Equation 4 by 3 to align the coefficients of y:

15x + 3y = 15000 (Equation 7)

Subtract Equation 7 from Equation 6:

(22x + 3y) - (15x + 3y) = 20250 - 15000
7x = 5250
x = 750

Substitute x = 750 back into Equation 4:

5(750) + y = 5000
3750 + y = 5000
y = 1250

Now substitute the values of x and y back into Equation 2 to find z:

z = 4x = 4(750) = 3000

Therefore, the amounts invested at each interest rate are:
2%: $750
3%: $1250
5%: $3000